Proving Trigonometric Identities - Intermediate

Proving Trigonometric Identities - Advanced

Proving Trigonometric Identities

Proving a trigonometric identity refers to showing that the identity is always true, no matter what value of \( x \) or \( \theta \) is used.

Because it has to hold true for all values of \(x\), we cannot simply substitute in a few values of \(x\) to "show" that they are equal. It is possible that both sides are equal at several values (namely when we solve the equation), and we might falsely think that we have a true identity.

Instead, we have to use logical steps to show that one side of the equation can be transformed to the other side of the equation. Sometimes, we will work separately on each side, till they meet in the middle.

General Approach

There are many different ways to prove an identity. Here are some guidelines in case you get stuck:

1) Work on the side that is more complicated. Try and simplify it.
2) Replace all trigonometric functions with just \( \sin \theta \) and \( \cos \theta \) where possible.
3) Identify algebraic operations like factoring, expanding, distributive property, adding and multiplying fractions. This allows us to simplify the expression further.
4) Use the various trigonometric identities. In particular, watch out for the Pythagorean identity.
5) Work from both sides.
6) Keep an eye on the other side, and work towards it.
7) Consider the "trigonometric conjugate."

Prove the identity

\[ \frac { \cot \theta } { \csc \theta } = \cos \theta. \]

(Guideline 1)
We start from the LHS which is more complicated. Notice that if we were to start from the RHS, it is not clear how we can proceed.

(Guideline 2 ... We might want to replace with \( \sin \theta \) and \( \cos \theta \) to proceed further. However, since everything is in just \( \sec \theta \) and \( \tan \theta \), let's keep it that way for now.)

We now have the \( \sin x - \cos x \) which is in the numerator of the LHS, so we know that the denominator of the RHS must be something similar. In fact, if this is an identity, it must be equal to \( (\sin x + \cos x)^ 2 \). We can verify that

At first it might be tempting to factorize \(\sin^2 A-\sin^2 B\) into \((\sin A+\sin B)(\sin A-\sin B)\), but you will soon realize that this leads to nothing. In order to move on, we should apply the power reduction formula, which gives

The solution above might be quite difficult to think of. If you find yourself stuck while struggling to organize the LHS, switching over and tackling the RHS might work. Just try converting the tangents into sines and cosines by using the formula \(\tan=\frac{\sin\theta}{\cos\theta}\), and solve the problem in the opposite direction of the first solution, as shown below: